When photons of energy 4.25 eV strike the surface of metal, A, the ejected photoelectrons have maximum kinetic energy $T_A$ and de Broglie wavelength ${\lambda }_A$.The maximum kinetic energy of  photoelectrons liberated from another metal B by photons of energy 4.70 eV is $T_B=T_A-1.50eV$ . If the de Broglie wavelength of these photoelectron is ${\lambda }_B=2{\lambda }_A\text{\hspace{0.17em},}$ then